Answer
$$\int^{1/4}_{1/6}2\cot\pi xdx=\frac{\ln2}{\pi}$$
Work Step by Step
$$A=\int^{1/4}_{1/6}2\cot\pi xdx=2\int^{1/4}_{1/6}\frac{\cos\pi x}{\sin \pi x}dx$$
Set $u=\sin\pi x$, which means $$du=\pi\cos\pi xdx$$ $$\cos\pi xdx=\frac{1}{\pi}du$$
- For $x=1/4$, we have $u=\sin\frac{\pi}{4}=\frac{\sqrt2}{2}$
- For $x=1/6$, we have $u=\sin\frac{\pi}{6}=\frac{1}{2}$
Therefore, $$A=\frac{2}{\pi}\int^{\sqrt2/2}_{1/2}\frac{1}{u}du$$ $$A=\frac{2}{\pi}\ln|u|\Big]^{\sqrt2/2}_{1/2}$$ $$A=\frac{2}{\pi}(\ln(\sqrt2/2)-\ln(1/2))$$ $$A=\frac{2}{\pi}\Big(\ln(2^{1/2}/2)-\ln1+\ln2\Big)$$ $$A=\frac{2}{\pi}\Big(\frac{1}{2}\ln2-\ln2+\ln2\Big)$$ $$A=\frac{2}{\pi}\Big(\frac{1}{2}\ln2\Big)=\frac{\ln2}{\pi}$$
